Math 533 Statistics: Course Project B

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Based on the statistical analysis, we can support the claim that the average (mean) annual income was less than $50,000. We cannot support the claim that the true population proportion of customers who live in an urban area exceeds 40%, the average number of years lived in the current home is less than 13 years and the average credit balance for suburban customers is more than $4,300. a. the average (mean) annual income was less than $50,000, | INCOME($1000) | count | 50 | mean | 43.48 | sample standard deviation | 14.55 | Ho: µ =50000 Ha: µ<50000 The test statistic is Z=(xbar-µ)/(s/√n) =(43480-50000)/(14550.7/√50) =-3.17 The p-value= P(Z<-3.17)= 0.0008 (from standard normal table). Since the p-value is less than 0.05, we reject Ho. So we can conclude that the average (mean) annual income was less than $50,000. Given a=0.05, the critical value is |Z(0.025)|=1.96 (from standard normal table) So 95% CI is xbar +/- Z*s/√n --> 43480 +/- 1.96*14550.7/√50 --> (39446.75, 47513.25) We have 95% confident that the population mean will be within this interval. Based on the statistical analysis, we can support the claim that the average (mean) annual income was less than $50,000. b. the true population proportion of customers who live in an urban area exceeds 40% The test hypothesis is Ho: p=0.4 Ha: p>0.4 The test statistic is Z=(phat-p)/√(p*(1-p)/n) =(21/50-0.4)/√(0.4*0.6/50) =0.29 The p-value= P(Z>0.29)= 0.3859 (from standard normal table) Since the p-value is larger than 0.05, we do not reject Ho. So we cannot conclude that the true population proportion of customers who live in an urban area exceeds 40% So 95% CI is phat +/- Z*√(p*(1-p)/n) --> 21/50 +/- 1.96*√(0.4*0.6/50) --> (0.2842072, 0.5557928) We have 95% confident that the population proportion will be within this interval.

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